A naked patient undergoes surgery in a warm room. Determine the ratio Pe/Pe. where Pe is the heat-power lost from the patient due to air convection and Pe is the heat-power lost from the patient due to evaporation. Take the skin temperature to be 32 °C, the room temperature to be 29°C, the total body area to be 1.3 m2, the emissivity e = 1.0, and the convection of coefficient, q = 7.1 WI(m2.K). Take water to be lost from the patient by evaporation at a rate of 40 g/(m?.hr) and assume this to be pure water. Take any other coefficients required from the course notes.

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A naked patient undergoes surgery in a warm room. Determine the ratio Pe/Pe.
where Pe is the heat-power lost from the patient due to air convection and Pe is the
heat-power lost from the patient due to evaporation.
Take the skin temperature to be 32 °C, the room temperature to be 29°C, the total
body area to be 1.3 m2, the emissivity e = 1.0, and the convection of coefficient, q =
7.1 WI(m?.K). Take water to be lost from the patient by evaporation at a rate of 40
g/(m?.hr) and assume this to be pure water. Take any other coefficients required from
the course notes.
1.1
1.3
1100
850
0.85
1300
Transcribed Image Text:A naked patient undergoes surgery in a warm room. Determine the ratio Pe/Pe. where Pe is the heat-power lost from the patient due to air convection and Pe is the heat-power lost from the patient due to evaporation. Take the skin temperature to be 32 °C, the room temperature to be 29°C, the total body area to be 1.3 m2, the emissivity e = 1.0, and the convection of coefficient, q = 7.1 WI(m?.K). Take water to be lost from the patient by evaporation at a rate of 40 g/(m?.hr) and assume this to be pure water. Take any other coefficients required from the course notes. 1.1 1.3 1100 850 0.85 1300
Expert Solution
Step 1

The skin temperature Ts=320C

The room temperature TR=290C

The total body area A=1.3 m2

The emissivity e=1

The convection coefficient q=7.1 W/m2K 

Heat lost from skin through convection

 HC=eqATs-TRHC=17.1 W/m2K1.3 m232°C-29°CHC=27.69 J

Step 2

The rate of  water evaporated from unit area of the skin mt=40 g/m2hr

The total amount of water evaporated in one hour

m=40×10-3kg/m2h1.3 m21 hr

The total amount of water evaporated in one second m=40×10-3 kg/m2 1.3 m213600 s=1.44×10-5 kg

Latent heat of vaporization of water Lf=2260×103 J/kg

Heat absorbed for evaporation of water from skin

 He=mLHe=1.44×10-5 kg2260×103 J/kgHe=32.64 J

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